Question: Simplify; express your answer in exponential form. Assume $a\neq 0, y\neq 0$. $\dfrac{{(a^{3})^{-2}}}{{(a^{3}y^{-4})^{-3}}}$
Answer: To start, try working on the numerator and the denominator independently. In the numerator, we have ${a^{3}}$ to the exponent ${-2}$ . Now ${3 \times -2 = -6}$ , so ${(a^{3})^{-2} = a^{-6}}$ In the denominator, we can use the distributive property of exponents. ${(a^{3}y^{-4})^{-3} = (a^{3})^{-3}(y^{-4})^{-3}}$ Simplify using the same method from the numerator and put the entire equation together. $\dfrac{{(a^{3})^{-2}}}{{(a^{3}y^{-4})^{-3}}} = \dfrac{{a^{-6}}}{{a^{-9}y^{12}}}$ Break up the equation by variable and simplify. $\dfrac{{a^{-6}}}{{a^{-9}y^{12}}} = \dfrac{{a^{-6}}}{{a^{-9}}} \cdot \dfrac{{1}}{{y^{12}}} = a^{{-6} - {(-9)}} \cdot y^{- {12}} = a^{3}y^{-12}$.